Answer
$[H_2] = 0.05 M$
$[CO_2] = 0.05 M$
$[H_2O] = 0.11 M$
$[CO] = 0.11 M$
Work Step by Step
- Calculate all the concentrations:
$$[H_2] = ( 0.80 )/(5.0) = 0.16 M$$
$$[CO_2] = ( 0.80 )/(5.0) = 0.16 M$$
1. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ H_2O ][ CO ]}{[ H_2 ][ CO_2 ]}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ H_2 ] = 0.16 \space M - x$
$ [ CO_2 ] = 0.16 \space M - x$
$ [ H_2O ] = x$
$ [ CO ] = x$
$$4.2 = \frac{(0 + x)(0 + x)}{(0.16 - x)(0.16 - x)}$$
x = 0.11 M
$[H_2] = 0.05 M$
$[CO_2] = 0.05 M$
$[H_2O] = 0.11 M$
$[CO] = 0.11 M$
